Question

我有一个带有字符串的列表清单：

list = [["a","b"],["c","d"],["a", "e"],["f","d"],["x","y"]]

现在我要合并所有列表，其中有1个类似项，如下所示：

grouped_list = [["a", "b", "e"],["c","d","f"],["x","y"]]

我的代码现在是这样：

 list = [["a","b"],["b","c"],["d","e"],["x","y"]]
 clist = list.copy()
 result = []
 counter = 0
 del_list = []
 def oneofsame(L1, L2):
     counter = 0
     for i in L1:
         for j in L2:
             if i == j:
                 counter += 1
     if counter == 0:
         return False
     else:
         return True

 for l in list:
     try:
         del clist[clist.index(l)]
     except:
         pass
     result.append([])
     for i in l:
         for cl in clist:
             if oneofsame(l, cl):
                 for j in l:
                     if j not in result[counter]:
                         result[counter].append(j)
                 for j in cl:
                     if j not in result[counter]:
                         result[counter].append(j)
                 del_list.append(cl)
             else:
                 result.append(cl)
                 del_list.append(cl)
             for j in del_list:
                 del clist[clist.index(j)]
             del_list = []     
     counter += 1

 del_list = []
 cresult = result.copy()
 for i in range(len(cresult)-1, 0, -1):
     if cresult[i] == []:
         del result[i]
 print(result)

但是此代码不会合并所有示例输入（我无法粘贴示例输入，因为它的敏感数据）

Answer 1

这是一种方法。

对于每对：

如果我们找到一个包含其中一个值的组，则将该对附加到该组上
如果找到另一个包含另一个值的组，则将这些组合并。
如果找不到匹配的组，那么我们的对就是一个新的组。

def group_equals(lst):
    groups = []

    for pair in lst:
        pair = set(pair)
        equals_found = 0
        for idx, group in enumerate(groups):
            if group.intersection(pair):
                equals_found += 1
                if equals_found == 1:
                    # We found a first group that contains one of our values,
                    # we can add our pair to the group
                    group.update(pair)
                    first_group = group
                elif equals_found == 2:
                    # We found a second group that contains the other one of 
                    # our values, we merge it with the first one
                    first_group.update(group)
                    del groups[idx]
                    break
        # If none of our values was found, we create a new group
        if not equals_found:
            groups.append(pair)

    return [list(sorted(group)) for group in groups]

tests = [ [["a", "b"], ["c", "d"], ["b", "c"]],  # all equal
          [["a","b"],["c","d"],["a", "e"],["f","d"]],
          [["a","b"],["c","d"],["a", "e"],["f","d"],["x","y"]]
        ]

for lst in tests:
    print(group_equals(lst))

# [['a', 'b', 'c', 'd']]
# [['a', 'b', 'e'], ['c', 'd', 'f']]
# [['a', 'b', 'e'], ['c', 'd', 'f'], ['x', 'y']]

Answer 2

希望下面的代码能解决您的问题：

import itertools
import copy

lista = [["a","b"],["c","d"],["a", "e"],["f","d"],["x","y"]] #[["a","b"],["e","d1"],["a", "e"],["a","d"],["d","y"]]

def grouped_list(lista):
    aa = []
    bbc = copy.deepcopy(lista)
    flag = False

    for a, b in itertools.combinations(lista,2):
        bb = a+b

        if len(set(bb)) < len(bb):
            flag = True
            cc = list(set(bb))
            cc.sort()

            if cc not in aa: aa.append(cc)
            if a in lista:   lista.remove(a) 
            if b in lista:   lista.remove(b)

    if lista:        aa = aa + lista
    if not flag:     return bbc
    else:            return grouped_list(aa)


print ("Grouped list -->", grouped_list(lista))

可以随意询问/建议上面的代码中的任何内容。

将列表合并到在python中具有一个相似项的列表中

2 个答案: